$$-\frac{e^t}{2}\int{e^{-t}\tanh{(t)}} dt$$
For the integrand ${e^{-t}\tanh{t}}$, write $\tanh{t}$ as $\frac{e^t-e^{-t}}{e^{-t}+e^t}$.
(Note that if this doesn't look familiar, you should review hyperbolic sine and cosine)
$$-\frac{e^t}{2}\int{e^{-t}\tanh{(t)}}dt=-\frac{e^t}{2}\int{\frac{e^{-t}(e^t-e^{-t})}{e^{-t}+e^t}}dt$$
Expanding the integrand gives
$$=-\frac{e^t}{2}\int{\left(\frac{e^t}{e^{2t}+1}-\frac{e^{-t}}{e^{2t}+1}\right)}dt=-\frac{e^t}{2}\int{\left(\frac{e^t}{e^{2t}+1}-\frac{e^{-t}}{e^{2t}+1}\right)}dt$$
For the integrand $\left(\frac{e^t}{e^{2t}+1}\right)$, substitute $u=e^x$ and $du=e^x dx$, and for the integrand $\left(\frac{e^{-t}}{e^{2t}+1}\right)$, substitute $v=e^x$ and $dv=e^x dx$.
Try proceeding this way. If you need help beyond this point, just lmk
